A 1,200-N crate is being pushed across a level floor at a constant speed by a force F of 310 N at an angle of 20.0° below the horizontal, as shown in the figure a below.20.0°<--120.0°(a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.)0.222(b) If the 310-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is thesame as that found in part (a).|m/s2

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Description: A 1,200-N crate is being pushed across a level floor at a constant speed by a force F of 310 N at an angle of 20.0° below the horizontal, as shown in the figure a below.20.0°<--120.0°(a) What is the coefficient of kinetic friction between the crate

Given data The weight of the crate W = 1200 N. The force is given as F = 310 N. The angle between…

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A 1,200-N crate is being pushed across a level floor at a constant speed by a force F of 310 N at an angle of 20.0° below the horizontal, as shown in the figure a below.
20.0°
<--120.0°
(a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.)
0.222
(b) If the 310-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is the
same as that found in part (a).
|m/s2

Expert Solution

Step 1

Given data

The weight of the crate W = 1200 N.
The force is given as F = 310 N.
The angle between the force and horizontal is given as $θ = 20 °$ .

a) The free-body diagram is drawn as,

Balancing the vertical force is given as,

$N = F \sin 20 ° + W$ ............(1)

Now, balancing the horizontal force is given as,

$\begin{array}{rcl} f & = & F \cos 20 ° \\ μ N & = & F \cos 20 ° . . . . . . . . . . . . . . (2) \end{array}$

From equation (1) and (2) as,

$\begin{array}{rcl} μ (F \sin 20 ° + W) & = & F \cos 20 ° \\ μ & = & \frac{F \cos 20 °}{(F \sin 20 ° + W)} \\ μ & = & \frac{310 N \times \cos 20 °}{(310 N \times \sin 20 ° + 1200 N)} \\ μ & = & 0.222 \end{array}$

Thus, the coefficient of friction is 0.222.

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